Question

A \(1.00-\mu \mathrm{F}\) capacitor charged to \(50.0 \mathrm{~V}\) and a \(2.00-\mu \mathrm{F}\) capacitor charged to \(20.0 \mathrm{~V}\) are connected, with the positive plate of each connected to the negative plate of the other. What is the final charge on the \(1.00-\mu \mathrm{F}\) capacitor?

Short answer

Answer: The final charge on the 1.00-µF capacitor is 135 µC.

Step-by-step solution

Identify the given quantities and the final goal

We are given: - 1st capacitor: C1 = 1.00 μF and V1 = 50.0 V - 2nd capacitor: C2 = 2.00 μF and V2 = 20.0 V Goal: Find the final charge (Q') on the 1.00-μF capacitor.

Calculate the initial charges on each capacitor

Use the formula Q = CV for each capacitor. Q1_initial = C1 * V1 = (1.00 μF)(50.0 V) = 50.0 µC Q2_initial = C2 * V2 = (2.00 μF)(20.0 V) = 40.0 µC

Find the total charge

Since neither capacitor loses or gains charge in the process, we can find the total charge by adding their initial charges. Q_total = Q1_initial + Q2_initial = 50.0 µC + 40.0 µC = 90.0 µC

Find the equivalent capacitance

Since the capacitors are connected in series, we can find the equivalent capacitance using the formula: 1 / C_eq = 1 / C1 + 1 / C2 Calculate C_eq: C_eq = (C1 * C2) / (C1 + C2) = (1.00 μF * 2.00 μF) / (1.00 μF + 2.00 μF) = 2/3 μF

Calculate the final common voltage

Using the formula Q = CV, we find the common voltage after charge redistribution. V_final = Q_total / C_eq = 90.0 µC / (2/3 μF) = 135 V

Calculate the final charge on the 1.00-µF capacitor

Use the formula Q = CV for the 1.00-µF capacitor and the final common voltage. Q1_final = C1 * V_final = (1.00 μF)(135 V) = 135 µC The final charge on the 1.00-µF capacitor is 135 µC.

Key concepts

Electrostatics

Electrostatics is a branch of physics that deals with the study of stationary or slow-moving electric charges. Charges can be either positive or negative and originate from the protons and electrons within atoms. When an object has an imbalance of these charges, it exhibits electrostatic properties, such as attracting or repelling other charged objects.

In the provided exercise, we explore a scenario involving two capacitors, each with an initial electric charge and voltage. When these capacitors are connected with opposing charges facing each other, they form an electrostatic system. The charges redistribute until a steady state is reached, showing electrostatic principles in action. In real-world applications, understanding electrostatic interactions is crucial for the design and function of electronic devices, such as capacitors in circuits.

Capacitance

Capacitance is a measure of a capacitor's ability to store electric charge per unit voltage. It is denoted by the symbol 'C' and is measured in farads (F). Capacitors, fundamental components in electronics, consist of two conductive plates separated by an insulating material, known as the dielectric.

In this exercise, we have two capacitors with known capacitances and voltages. When capacitors connect in series, their combined ability to store charge changes. The equivalent capacitance of a series circuit is calculated by taking the reciprocal sum of the individual capacitances: \[\frac{1}{{C_{eq}}} = \frac{1}{{C_1}} + \frac{1}{{C_2}}\].
The process of charge redistribution affects the final charge on each capacitor, which is directly proportional to its capacitance, as shown in the solutions steps for our exercise.

Electric Charge

Electric charge, represented by 'Q', is a fundamental property of matter carried by protons and electrons. It exists in discrete amounts and can be quantified as an excess or shortage of electrons. Charge is measured in coulombs (C), with the elementary charge of one electron being approximately \(-1.6 \times 10^{-19}\) coulombs.

In the context of our exercise, each capacitor initially possesses a specific charge determined by multiplication of its capacitance and voltage (\(Q = CV\)). Following their connection in series, the total charge remains conserved; it only redistributes between the capacitors. Through formula manipulation and understanding the properties of series connections, we derive the final charge on each capacitor, a core concept to grasp for any student studying electrostatics and electric circuit behavior.